What are the steps to configure your router in order?

Answer:

C. WordArt

Explanation:

WordArt feature of MS word is used for styling the text n order to make it look neater or more noticeable. These features are found in the top home menu. They feature text size, text style, text color, and options to make text bold, italic, underlined, etc. There are also many features to make text flashy, colorful, and add special effects. It can be shaded, neon, have shadows, etc.

When the process is in control but does not meet specification, it is referred to as a special cause error.

In statistical process control, a process is considered to be in control when it operates within the defined limits and shows only random variations. However, when a process is in control but does not meet the desired specifications, it indicates the presence of a special cause error.

Special cause errors are attributed to specific factors or events that cause the process to deviate from the expected outcome. These errors are typically unpredictable and require investigation and corrective action to bring the process back within the desired specifications.

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Answer:  0.765 m kg / s

Explanation:

Momentum = (mass) x (speed)

Answer:

Can't help

Explanation:

As a rule of thumb, the estimated depth of standard corrugated steel roof decking can be determined based on the span of the decking. For a span of 10 feet, the recommended depth is typically around 1.5 inches to 3 inches.

The depth of the corrugated steel roof decking is an important factor in determining its load-bearing capacity and overall structural stability. A deeper profile provides greater strength and rigidity, allowing the decking to support heavier loads and resist deflection.

However, it's important to note that this rule of thumb is a general guideline and may vary depending on specific design requirements, local building codes, and the intended use of the roof decking. It is always advisable to consult with structural engineers or professionals in the construction industry to determine the appropriate depth of corrugated steel roof decking for a specific project.

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Answer:

I think its A

Explanation:

Answer:

I am not sure it's confusing

(a) The coefficient of friction for the journal bearing is 0.018.

(b) The bearing pressure is 9.62 MPa.

(c) The heat generated is 183.6 W.

(d) The heat dissipated is 183.6 W.

(e) The grade of oil to be used is determined based on the viscosity temperature characteristics of the oil.

(f) Artificial cooling is not required for an unventilated, average industrial bearing.

(a) The coefficient of friction for a journal bearing can be calculated using the equation:

μ = ZN / (π x L x d x n)

Where μ is the coefficient of friction, ZN is the viscosity of the oil, L is the length of the bearing, d is the diameter of the bearing, and n is the rotational speed. Plugging in the given values, we get:

μ = \((30 x 10^-3)\)/ (π x 0.15 x 0.1 x 2000) = 0.018

(b) The bearing pressure can be calculated using the equation:

P = F / (π x L x d)

Where P is the bearing pressure and F is the radial load. Plugging in the given values, we get:

P = 43,000 N / (π x 0.15 x 0.1) = 9.62 MPa

(c) The heat generated in the bearing can be calculated using the equation:

Q = F x μ x d x n

Where Q is the heat generated and the other variables are as defined earlier. Plugging in the given values, we get:

Q = 43,000 N x 0.018 x 0.1 x 2000 = 183.6 W

(d) The heat dissipated from the bearing is equal to the heat generated since it is assumed that there is no heat transfer to the surroundings.

(e) The grade of oil to be used depends on the viscosity-temperature characteristics of the oil. The specific grade can be determined by referring to oil viscosity-temperature charts provided by oil manufacturers.

(f) Artificial cooling is not required for an unventilated, average industrial bearing since the heat generated is equal to the heat dissipated. However, if the heat generated exceeds the heat dissipated, artificial cooling methods such as cooling fins or forced air circulation may be necessary.

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Answer:

Gauge Pressure = 408.3 KPa

Explanation:

The pressure inside the pipe can be given in terms of the elevation, by the following formula:

P = ρgΔz

where,

P = Absolute Pressure = ?

ρ = Density of Water = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

Δz = elevation = 52 m

Therefore,

P = (1000 kg/m³)(9.8 m/s²)(52 m)

P = 509.6 KPa

Now, for gauge pressure:

Gauge Pressure = P - Atmospheric Pressure

Gauge Pressure = 509.6 KPa - 101.3 KPa

Gauge Pressure = 408.3 KPa

Answer:

the welding gun liner regulates the shielding gas.

Explanation:

The purpose of the welding gun liner is to properly position the welding wire from the wire feeder till it gets to the nozzle or contact tip of the gun. Regulation of the shielding gas depends on factors such as the speed, current, and type of gas being used. In gas metal arc welding, an electric arc is used to generate heat which melts both the electrode and the workpiece or base metal.

The electric arc produced is shielded from contamination by the shielding gas. The heat generated by the short electric arc is low.

Answer:

When a system is produced using the evolutionary development model, features tend to be added without regard to an overriding design. With each modification, the software becomes increasingly disorganized. System maintenance hampered by these problems, as it is harder identifying the source of bugs in poorly designed systems. Also, keeping the documentation up to date over successive "evolution" is uncommon. Poor documentation also makes maintenance more difficult.

Explanation:

・It leads to implementing and then repairing way of building systems.

・Practically, this methodology may increase the complexity of the system as scope of the system may expand beyond original plans.

・Incomplete application may cause application not to be used as the full system was designed.

・Their results have incomplete or inadequate problem analysis.

Answer:

In my opinion, Energy, Time, Money and Development

Answer:

Pollution, time, money, and energy.

Explanation:

The rate of change of the radius with time is a partial derivative of the rate of change of volume with time. At the instant the height of the cylinder is 6 feet is the radius is decreasing at a rate of approximately 4.634 ft./s.

The height of the cone is 27 meters when the radius of the cone is 55 meters and the volume of the cone is 733 cubic meters.The radius of a cylinder is increasing at a constant rate of 3 meters per second, and the volume is increasing at a rate of 108 cubic meters per second.

Cones grow in size at constant rates of 9 meters per second for the radius and 1631 cubic meters per second for the volume.

Therefore, The height of the cone is 27 meters when the radius of the cone is 55 meters and the volume of the cone is 733 cubic meters.

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One environmental disadvantage of nuclear energy is the risk of radiation leaks, which can cause long-term damage to the environment and human health.

Nuclear energy has many advantages, such as being a clean and efficient source of energy that does not produce carbon emissions. However, there are also some environmental disadvantages associated with it. One of the most significant is the risk of radiation leaks, which can cause serious damage to the environment and human health. Radiation leaks can occur due to equipment failure, natural disasters, or human error during the production or storage of nuclear materials.

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Answer:

Top Final year projects for civil engineering students

• Geographic Information System using Q-GIS. ...

• Structural and Foundation Analysis. ...

• Construction Project Management & Building Information Modeling. ...

• Tall Building Design. ...

• Seismic Design using SAP2000 & ETABS.

Explanation:

Hope it's help

A nutrunner on the engine assembly line has been failing for low torque. process includes identifying the root cause of the fault, and optimizing the nut runner's performance.

The first step would be to investigate the cause of the low torque issue in the nut runner. This may involve examining the machine, reviewing maintenance records, and gathering data on when and how the fault occurs. Once the root cause is identified, corrective actions can be taken. This may include repairing or replacing faulty components, recalibrating the nut runner, or updating software/firmware.

To prevent future occurrences, implementing a preventive maintenance program is crucial. Regular inspections, scheduled maintenance tasks, and performance testing can help identify and address potential issues before they lead to line stoppages. Additionally, providing thorough training to operators and maintenance staff on nutrunner operation, maintenance procedures, and troubleshooting techniques can contribute to quicker resolution of faults and reduce downtime.

Continuous monitoring of the nutrunner's performance is essential to ensure it operates within specified tolerances. This can be done through real-time data collection and analysis, including torque measurement and trend analysis. By closely monitoring the nutrunner's performance, any deviations or anomalies can be detected early, allowing for proactive interventions.

Overall, a systematic approach that combines investigation, preventive maintenance, employee training, and continuous monitoring can help solve the problem of the faulty nutrunner and improve the efficiency and productivity of the assembly line.

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Temperature is a critical aspect of our lives as it governs our behavior and the natural world around us. Celsius and Fahrenheit are the two temperature scales that are used all over the world.

Fahrenheit to Celsius conversion is possible by using the formula C= (5/9) x (F-32) and

Celsius to Fahrenheit conversion can be done by using the formula F= (9/5) x C + 32.

Below are the steps to create an application that allows the user to enter a temperature:

Step 1: Open the Visual Studio IDE and create a new project of type Windows Forms App (.NET Framework). Name the project CelsiusToFahrenheitConversion.

Step 2: From the Toolbox, drag two TextBox controls and place them on the form. Name the TextBox controls txtCelsius and txtFahrenheit.

Step 3: From the Toolbox, drag two Button controls and place them on the form. Name the Button controls btn Convert Celsius To Fahrenheit and btn Convert Fahrenheit To Celsius.

Step 4: Double-click the Convert to Fahrenheit button. Write the code to convert the Celsius temperature entered in the txtCelsius TextBox control to Fahrenheit, and then display the result in the txtFahrenheit TextBox control. The code is as follows:

```private void btnConvertCelsiusToFahrenheit_Click(object sender, EventArgs e)
{
   double celsius = double.Parse(txtCelsius.Text);
   double fahrenheit = (9.0 / 5.0) * celsius + 32.0;
   txtFahrenheit.Text = fahrenheit.ToString();
}```

Step 5: Double-click the Convert to Celsius button. Write the code to convert the Fahrenheit temperature entered in the txtFahrenheit TextBox control to Celsius, and then display the result in the txtCelsius TextBox control. The code is as follows:

```private void btnConvertFahrenheitToCelsius_Click(object sender, EventArgs e)
{
   double fahrenheit = double.Parse(txtFahrenheit.Text);
   double celsius = (5.0 / 9.0) * (fahrenheit - 32.0);
   txtCelsius.Text = celsius.ToString();
}```

Step 6: Save and run the application. Now you can enter a temperature in either Celsius or Fahrenheit, and click the corresponding button to convert it to the other scale.

In conclusion, Celsius and Fahrenheit Temperature Converter application is a handy application that can be used to convert temperature from Celsius to Fahrenheit and Fahrenheit to Celsius.

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A process generates 250 watts of useful energy and 600 watts of waste energy, the efficiency of the process is 29%.

To perform a task or achieve a goal while using less energy is known as energy efficiency. Energy-efficient homes, buildings, manufacturing facilities, and electronics use less energy to produce goods and less energy to heat, cool, and operate appliances and electronics.

One of the simplest and most affordable strategies for halting climate change, lowering consumer energy costs, and boosting the competitiveness of American companies is energy efficiency. Achieving net-zero carbon dioxide emissions through decarbonization depends on energy efficiency as well.

The Office of Energy Efficiency and Renewable Energy (EERE) supports clean energy through its technical offices and programs that support research and development and encourage energy efficiency in all facets of the American economy.

Energy efficiency = useful energy output/energy input × 100%

useful energy output = 250 watts

energy input = useful energy output  + waste energy ouputs

                      = 250 + 600

                       = 850 watts

Energy efficiency = (250 /850) × 100%

                              = 29%

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Answer:

\(D=0.1160m\)

Explanation:

From the question we are told that:

Output Power \(P=1.2kw\)

Density \(\rho=1.29kg/m^3\)

Wind speed \(V=17.5mph=>7.8m/s\)

Efficiency \(\gamma=60\%=>0.60\)

Let Betz Limit

  \(C_p=\frac{16}{27}\)

Generally the equation for Turbine Efficiency is mathematically given by

  \(\gamma=\frac{P}{P'}\)

Where

P'=input power

 \(P' = In\ power\)

 \(P'=1/2*C_p*\rho u^3*A\)

 \(P'=1/2*C_p*\rho u^3*\frac{\pi}{4}*D^2\)

Therefore the blade diameter for a two-blade propeller type rotor is

 \(\gamma=\frac{P*2*27*4}{16*\rho u^3*\pi*D^2}\)

 \(D^2=\frac{P*2*27*4}{16*\rho u^3*\pi*\gamma}\)

 \(D^2=\frac{1.2*2*27*4}{16*1.29*7.91^3*\pi*0.60}\)

 \(D^2=0.0135\)

 \(D=\sqrt{0.0135}\)

 \(D=0.1160m\)

Answer:

0.1160

Explanation: give them brainliest they deserve it

Answer: yayyyy free sh.it

Explanation:

Answer:

b.9

Explanation:

not sure

Answer:

the correct answer for the question is A. Agonizer. David is an agonizer as he agonizes while making decisions.

Explanation:

Machined parts can be classified as rotational or nonrotational. The two examples of operations that create nonrotational geometries are option (a) boring and option (d) planing.

Boring is a machining process that enlarges an existing hole to a desired diameter, and the resulting geometry is a straight cylindrical shape with parallel walls. Planing, on the other hand, is a process that produces flat surfaces by removing material in a linear motion. The resulting geometry is a flat surface with straight edges.

Drilling and turning are examples of operations that create rotational geometries. Drilling involves creating a hole by removing material in a rotary motion, resulting in a cylindrical shape with a circular cross-section. Turning, on the other hand, is a process that creates round geometries by removing material with a rotating cutting tool.

Milling is an operation that can create both rotational and nonrotational geometries, depending on the type of milling process used. Some milling processes involve rotating the workpiece, resulting in rotational geometries, while others involve moving the cutting tool in a linear motion, resulting in nonrotational geometries.

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Answer:

7 available

Explanation:

Since 3 colors are available r = 3

Total combination = 35

nCr = 35 ---1

nCr = n!/(n-r)!r!---2

We put equation 1 and 2 together

n-1)(n-2)(n-3)!/n-3)! = 35x 3!

We cancel out (n-3)!

(n-1)(n-2) = 210

7x6x5 = 210

nC3 = 35

7C3 = 35

So If there are 35 combinations, 7 colors are available.

Thank you!

The available colors are illustrations of combination, and there are 7 colors available.

The given parameters are:

\(Total = 35\) --- the number of combinations

\(r = 3\) -- the number of colors in one combination.

The number of combinations is calculated using:

\(Total = ^nC_r\)

Where n represents the number of colors available.

So, we have:

\(35= ^nC_3\)

Apply combination formula

\(35= \frac{n!}{(n - 3)!3!}\)

Expand the numerator

\(35= \frac{n(n - 1)(n - 2)(n - 3)!}{(n - 3)!3!}\)

Cancel out the common factors

\(35= \frac{n(n - 1)(n - 2)}{3!}\)

Expand

\(35= \frac{n(n - 1)(n - 2)}{3 \times 2 \times 1}\)

\(35= \frac{n(n - 1)(n - 2)}{6}\)

Multiply both sides by 6

\(210= n(n - 1)(n - 2)\)

Rewrite the equation as

\(n(n - 1)(n - 2) = 210\)

Using a graphing calculator, we have:

\(n = 7\)

Hence, the number of colors available is 7

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Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= \(density\times g\times \frac{h}{2}\)

By putting values, we get

= \(1000\times 9.8\times \frac{12.2}{2}\)

= \(1000\times 9.8\times 6.1\)

= \(59780\)

hence,

The average force will be:

= \(Pressure\times Area\)

= \(59780\times 3.6\times 12.2\)

= \(2625537 \ N\)

Or,

= \(2625 \ kN\)

Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V\(_s\)/( 1 - D )

given that; V\(_s\) = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

\(I_L\) = V\(_s\) / ( 1 - D )²R

given that R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6

we substitute

\(I_L\) = 20 / (( 1 - 0.6 )² × 12.5)

\(I_L\) = 20 / (( 0.4)² × 12.5)

\(I_L\) = 20 / ( 0.16 × 12.5 )

\(I_L\) = 20 / 2

\(I_L\) = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

\(I_{Lmax\) = [V\(_s\) / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

\(I_{Lmax\) = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

\(I_{Lmax\) = [20 / 2 ] + [ 60 / 20 ]    

\(I_{Lmax\) = 10 + 3

\(I_{Lmax\) = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

\(I_{Lmax\) = [V\(_s\) / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V\(_s\) = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

\(I_{Lmin\) = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

\(I_{Lmin\) = [20 / 2 ] -[ 60 / 20 ]    

\(I_{Lmin\) = 10 - 3

\(I_{Lmin\)  = 7 A

Therefore, the maximum inductor current is 7 A

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

\(I_D\) = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

\(I_D\) = 50 / 12.5

\(I_D\) = 4 A

Therefore, the average current in the diode under ideal components is 4 A

The value of the site is $10,472,000.

The total construction costs are:

$6,000/m2 * 10,000m2 + $400,000 = $64,000,000

The professional fees are:

$64,000,000 * 0.13 = $8,320,000

The total development costs are:

$64,000,000 + $8,320,000 = $72,320,000

The developer's return for risk and profit is:

$72,320,000 * 0.20 = $14,464,000

The total cost of the development is:

$72,320,000 + $14,464,000 = $86,784,000

The expected rent is:

$3,000/m2 * 8,000m2 * 12 months = $28,800,000

The initial yield is:

$28,800,000 * 0.07 = $2,016,000

The value of the site is:

$86,784,000 - $2,016,000 = $10,472,000

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8.76 MUs of energy will be produced yearly by the micro-hydro dam.

A dam is a physical obstruction that slows or restricts the flow of subsurface or seawater. Dam-created reservoirs offer freshwater for uses.

Taking into account that the micro-hydro dam produces 1 MW of power

To estimate how much energy this will yearly generate.

There is one unit of electricity per kwh ( kW hours)

A 1000 KW plant becomes a 1 MW plant.

The quantity of toward the in a day when a plant runs continuously at maximum capacity is:

24000 Kwh, approximately 24000 units, is equal to 1000 KW multiplied by 24 hours. Units made during a specific year.

When 1000 KW is multiplied by 24 hours and 365 days, the result is 8,760,000 Kwh, or 8.76 Million Units (MUs).

The micro-hydro dam will generate 8.76 MUs of energy every year.

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Answer:

Explanation:

"Safety helmet" redirects here. It is not to be confused with hard hat.

Drug Enforcement Administration (DEA) agents wearing Level B hazmat suits

Personal protective equipment (PPE) is protective clothing, helmets, goggles, or other garments or equipment designed to protect the wearer's body from injury or infection. The hazards addressed by protective equipment include physical, electrical, heat, chemicals, biohazards, and airborne particulate matter. Protective equipment may be worn for job-related occupational safety and health purposes, as well as for sports and other recreational activities. "Protective clothing" is applied to traditional categories of clothing, and "protective gear" applies to items such as pads, guards, shields, or masks, and others. PPE suits can be similar in appearance to a cleanroom suit.

The purpose of personal protective equipment is to reduce employee exposure to hazards when engineering controls and administrative controls are not feasible or effective to reduce these risks to acceptable levels. PPE is needed when there are hazards present. PPE has the serious limitation that it does not eliminate the hazard at the source and may result in employees being exposed to the hazard if the equipment fails.[1]

Any item of PPE imposes a barrier between the wearer/user and the working environment. This can create additional strains on the wearer; impair their ability to carry out their work and create significant levels of discomfort. Any of these can discourage wearers from using PPE correctly, therefore placing them at risk of injury, ill-health or, under extreme circumstances, death. Good ergonomic design can help to minimise these barriers and can therefore help to ensure safe and healthy working conditions through the correct use of PPE.

Practices of occupational safety and health can use hazard controls and interventions to mitigate workplace hazards, which pose a threat to the safety and quality of life of workers. The hierarchy of hazard controls provides a policy framework which ranks the types of hazard controls in terms of absolute risk reduction. At the top of the hierarchy are elimination and substitution, which remove the hazard entirely or replace the hazard with a safer alternative. If elimination or substitution measures cannot apply, engineering controls and administrative controls, which seek to design safer mechanisms and coach safer human behavior, are implemented. Personal protective equipment ranks last on the hierarchy of controls, as the workers are regularly exposed to the hazard, with a barrier of protection. The hierarchy of controls is important in acknowledging that, while personal protective equipment has tremendous utility, it is not the desired mechanism of control in terms of worker safety.rly PPE such as body armor, boots and gloves focused on protecting the wearer's body from physical injury. The plague doctors of sixteenth-century Europe also wore protective uniforms consisting of a full-length gown, helmet, glass eye coverings, gloves and boots (see Plague doctor costume) to prevent contagion when dealing with plague victims. These were made of thick material which was then covered in wax to make it water-resistant. A mask with a beak-like structure which was filled with pleasant-smelling flowers, herbs and spices to prevent the spread of miasma, the prescientific belief of bad smells which spread disease through the air.[2] In more recent years, scientific personal protective equipment is generally believed to have begun with the cloth facemasks promoted by Wu Lien-teh in the 1910–11 Manchurian pneumonic plague outbreak, although many Western medics doubted the efficacy of facemasks in preventing the spread of disease.[3]

Types

Personal protective equipment can be categorized by the area of the body protected, by the types of hazard, and by the type of garment or accessory. A single item, for example boots, may provide multiple forms of protection: a steel toe cap and steel insoles for protection of the feet from crushing or puncture injuries, impervious rubber and lining for protection from water and chemicals, high reflectivity and heat resistance for protection from radiant heat, and high electrical resistivity for protection from electric shock. The protective attributes of each piece of equipment must be compared with the hazards expected to be found in the workplace. More breathable types of personal protective equipment may not lead to more contamination but do result in greater user satisfaction.[4]

The use of a faulty PPE could be just as dangerous as not using any PPE at all because the user is still exposed to potential hazards and harm.

PPE is an acronym for personal protective equipment and it can be defined as a terminology that is used to denote any piece of equipment which offer protection to different parts of the body while working in a potentially hazardous environment.

Some examples of personal protective equipment (PPE) used to protect the different parts of the body are:

According to OSHA, the use of a faulty PPE could be just as dangerous as not using any PPE at all because the user is offered little or no protection at all.

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